It seems to be common knowledge that hash tables can achieve O(1), but that has never made sense to me. Can someone please explain it? Here are two situations that come to mind:
A. The value is an int smaller than the size of the hash table. Therefore, the value is its own hash, so there is no hash table. But if there was, it would be O(1) and still be inefficient.
B. You have to calculate a hash of the value. In this situation, the order is O(n) for the size of the data being looked up. The lookup might be O(1) after you do O(n) work, but that still comes out to O(n) in my eyes.
And unless you have a perfect hash or a large hash table, there are probably several items per bucket. So, it devolves into a small linear search at some point anyway.
I think hash tables are awesome, but I do not get the O(1) designation unless it is just supposed to be theoretical.
Wikipedia's article for hash tables consistently references constant lookup time and totally ignores the cost of the hash function. Is that really a fair measure?
Edit: To summarize what I learned:
It is technically true because the hash function is not required to use all the information in the key and so could be constant time, and because a large enough table can bring collisions down to near constant time.
It is true in practice because over time it just works out as long as the hash function and table size are chosen to minimize collisions, even though that often means not using a constant time hash function.
You have two variables here, m and n, where m is the length of the input and n is the number of items in the hash.
The O(1) lookup performance claim makes at least two assumptions:
Your objects can be equality compared in O(1) time.
There will be few hash collisions.
If your objects are variable size and an equality check requires looking at all bits then performance will become O(m). The hash function however does not have to be O(m) - it can be O(1). Unlike a cryptographic hash, a hash function for use in a dictionary does not have to look at every bit in the input in order to calculate the hash. Implementations are free to look at only a fixed number of bits.
For sufficiently many items the number of items will become greater than the number of possible hashes and then you will get collisions causing the performance rise above O(1), for example O(n) for a simple linked list traversal (or O(n*m) if both assumptions are false).
In practice though the O(1) claim while technically false, is approximately true for many real world situations, and in particular those situations where the above assumptions hold.
You have to calculate the hash, so the order is O(n) for the size of the data being looked up. The lookup might be O(1) after you do O(n) work, but that still comes out to O(n) in my eyes.
What? To hash a single element takes constant time. Why would it be anything else? If you're inserting n elements, then yes, you have to compute n hashes, and that takes linear time... to look an element up, you compute a single hash of what you're looking for, then find the appropriate bucket with that. You don't re-compute the hashes of everything that's already in the hash table.
And unless you have a perfect hash or a large hash table there are probably several items per bucket so it devolves into a small linear search at some point anyway.
Not necessarily. The buckets don't necessarily have to be lists or arrays, they can be any container type, such as a balanced BST. That means O(log n) worst case. But this is why it's important to choose a good hashing function to avoid putting too many elements into one bucket. As KennyTM pointed out, on average, you will still get O(1) time, even if occasionally you have to dig through a bucket.
The trade off of hash tables is of course the space complexity. You're trading space for time, which seems to be the usual case in computing science.
You mention using strings as keys in one of your other comments. You're concerned about the amount of time it takes to compute the hash of a string, because it consists of several chars? As someone else pointed out again, you don't necessarily need to look at all the chars to compute the hash, although it might produce a better hash if you did. In that case, if there are on average m chars in your key, and you used all of them to compute your hash, then I suppose you're right, that lookups would take O(m). If m >> n then you might have a problem. You'd probably be better off with a BST in that case. Or choose a cheaper hashing function.
The hash is fixed size - looking up the appropriate hash bucket is a fixed cost operation. This means that it is O(1).
Calculating the hash does not have to be a particularly expensive operation - we're not talking cryptographic hash functions here. But that's by the by. The hash function calculation itself does not depend on the number n of elements; while it might depend on the size of the data in an element, this is not what n refers to. So the calculation of the hash does not depend on n and is also O(1).
Hashing is O(1) only if there are only constant number of keys in the table and some other assumptions are made. But in such cases it has advantage.
If your key has an n-bit representation, your hash function can use 1, 2, ... n of these bits. Thinking about a hash function that uses 1 bit. Evaluation is O(1) for sure. But you are only partitioning the key space into 2. So you are mapping as many as 2^(n-1) keys into the same bin. using BST search this takes up to n-1 steps to locate a particular key if nearly full.
You can extend this to see that if your hash function uses K bits your bin size is 2^(n-k).
so K-bit hash function ==> no more than 2^K effective bins ==> up to 2^(n-K) n-bit keys per bin ==> (n-K) steps (BST) to resolve collisions. Actually most hash functions are much less "effective" and need/use more than K bits to produce 2^k bins. So even this is optimistic.
You can view it this way -- you will need ~n steps to be able to uniquely distinguish a pair of keys of n bits in the worst case. There is really no way to get around this information theory limit, hash table or not.
However, this is NOT how/when you use hash table!
The complexity analysis assumes that for n-bit keys, you could have O(2^n) keys in the table (e.g. 1/4 of all possible keys). But most if not all of the time we use hash table, we only have a constant number of the n-bit keys in the table. If you only want a constant number of keys in the table, say C is your maximum number, then you could form a hash table of O(C) bins, that guarantees expected constant collision (with a good hash function); and a hash function using ~logC of the n bits in the key. Then every query is O(logC) = O(1). This is how people claim "hash table access is O(1)"/
There are a couple of catches here -- first, saying you don't need all the bits may only be a billing trick. First you cannot really pass the key value to the hash function, because that would be moving n bits in the memory which is O(n). So you need to do e.g. a reference passing. But you still need to store it somewhere already which was an O(n) operation; you just don't bill it to the hashing; you overall computation task cannot avoid this. Second, you do the hashing, find the bin, and found more than 1 keys; your cost depends on your resolution method -- if you do comparison based (BST or List), you will have O(n) operation (recall key is n-bit); if you do 2nd hash, well, you have the same issue if 2nd hash has collision. So O(1) is not 100% guaranteed unless you have no collision (you can improve the chance by having a table with more bins than keys, but still).
Consider the alternative, e.g. BST, in this case. there are C keys, so a balanced BST will be O(logC) in depth, so a search takes O(logC) steps. However the comparison in this case would be an O(n) operation ... so it appears hashing is a better choice in this case.
TL;DR: Hash tables guarantee O(1) expected worst case time if you pick your hash function uniformly at random from a universal family of hash functions. Expected worst case is not the same as average case.
Disclaimer: I don't formally prove hash tables are O(1), for that have a look at this video from coursera [1]. I also don't discuss the amortized aspects of hash tables. That is orthogonal to the discussion about hashing and collisions.
I see a surprisingly great deal of confusion around this topic in other answers and comments, and will try to rectify some of them in this long answer.
Reasoning about worst case
There are different types of worst case analysis. The analysis that most answers have made here so far is not worst case, but rather average case [2]. Average case analysis tends to be more practical. Maybe your algorithm has one bad worst case input, but actually works well for all other possible inputs. Bottomline is your runtime depends on the dataset you're running on.
Consider the following pseudocode of the get method of a hash table. Here I'm assuming we handle collision by chaining, so each entry of the table is a linked list of (key,value) pairs. We also assume the number of buckets m is fixed but is O(n), where n is the number of elements in the input.
function get(a: Table with m buckets, k: Key being looked up)
bucket <- compute hash(k) modulo m
for each (key,value) in a[bucket]
return value if k == key
return not_found
As other answers have pointed out, this runs in average O(1) and worst case O(n). We can make a little sketch of a proof by challenge here. The challenge goes as follows:
(1) You give your hash table algorithm to an adversary.
(2) The adversary can study it and prepare as long as he wants.
(3) Finally the adversary gives you an input of size n for you to insert in your table.
The question is: how fast is your hash table on the adversary input?
From step (1) the adversary knows your hash function; during step (2) the adversary can craft a list of n elements with the same hash modulo m, by e.g. randomly computing the hash of a bunch of elements; and then in (3) they can give you that list. But lo and behold, since all n elements hash to the same bucket, your algorithm will take O(n) time to traverse the linked list in that bucket. No matter how many times we retry the challenge, the adversary always wins, and that's how bad your algorithm is, worst case O(n).
How come hashing is O(1)?
What threw us off in the previous challenge was that the adversary knew our hash function very well, and could use that knowledge to craft the worst possible input.
What if instead of always using one fixed hash function, we actually had a set of hash functions, H, that the algorithm can randomly choose from at runtime? In case you're curious, H is called a universal family of hash functions [3]. Alright, let's try adding some randomness to this.
First suppose our hash table also includes a seed r, and r is assigned to a random number at construction time. We assign it once and then it's fixed for that hash table instance. Now let's revisit our pseudocode.
function get(a: Table with m buckets and seed r, k: Key being looked up)
rHash <- H[r]
bucket <- compute rHash(k) modulo m
for each (key,value) in a[bucket]
return value if k == key
return not_found
If we try the challenge one more time: from step (1) the adversary can know all the hash functions we have in H, but now the specific hash function we use depends on r. The value of r is private to our structure, the adversary cannot inspect it at runtime, nor predict it ahead of time, so he can't concoct a list that's always bad for us. Let's assume that in step (2) the adversary chooses one function hash in H at random, he then crafts a list of n collisions under hash modulo m, and sends that for step (3), crossing fingers that at runtime H[r] will be the same hash they chose.
This is a serious bet for the adversary, the list he crafted collides under hash, but will just be a random input under any other hash function in H. If he wins this bet our run time will be worst case O(n) like before, but if he loses then well we're just being given a random input which takes the average O(1) time. And indeed most times the adversary will lose, he wins only once every |H| challenges, and we can make |H| be very large.
Contrast this result to the previous algorithm where the adversary always won the challenge. Handwaving here a bit, but since most times the adversary will fail, and this is true for all possible strategies the adversary can try, it follows that although the worst case is O(n), the expected worst case is in fact O(1).
Again, this is not a formal proof. The guarantee we get from this expected worst case analysis is that our run time is now independent of any specific input. This is a truly random guarantee, as opposed to the average case analysis where we showed a motivated adversary could easily craft bad inputs.
TL-DR; usually hash() is O(m) where m is length of a key
My three cents.
24 years ago when Sun released jdk 1.2 they fixed a bug in String.hashCode() so instead of computing a hash only based on some portion of a string since jdk1.2 it reads every single character of a string instead. This change was intentional and IHMO very wise.
In most languages builtin hash works similar. It process the whole object to compute a hash because keys are usually small while collisions can cause serious issues.
There are a lot of theoretical arguments confirming and denying the O(1) hash lookup cost. A lot of them are reasonable and educative.
Let us skip the theory and do some experiment instead:
import timeit
samples = [tuple("LetsHaveSomeFun!")] # better see for tuples
# samples = ["LetsHaveSomeFun!"] # hash for string is much faster. Increase sample size to see
for _ in range(25 if isinstance(samples[0], str) else 20):
samples.append(samples[-1] * 2)
empty = {}
for i, s in enumerate(samples):
t = timeit.timeit(lambda: s in empty, number=2000)
print(f"{i}. For element of length {len(s)} it took {t:0.3f} time to lookup in empty hashmap")
When I run it I get:
0. For element of length 16 it took 0.000 time to lookup in empty hashmap
1. For element of length 32 it took 0.000 time to lookup in empty hashmap
2. For element of length 64 it took 0.001 time to lookup in empty hashmap
3. For element of length 128 it took 0.001 time to lookup in empty hashmap
4. For element of length 256 it took 0.002 time to lookup in empty hashmap
5. For element of length 512 it took 0.003 time to lookup in empty hashmap
6. For element of length 1024 it took 0.006 time to lookup in empty hashmap
7. For element of length 2048 it took 0.012 time to lookup in empty hashmap
8. For element of length 4096 it took 0.025 time to lookup in empty hashmap
9. For element of length 8192 it took 0.048 time to lookup in empty hashmap
10. For element of length 16384 it took 0.094 time to lookup in empty hashmap
11. For element of length 32768 it took 0.184 time to lookup in empty hashmap
12. For element of length 65536 it took 0.368 time to lookup in empty hashmap
13. For element of length 131072 it took 0.743 time to lookup in empty hashmap
14. For element of length 262144 it took 1.490 time to lookup in empty hashmap
15. For element of length 524288 it took 2.900 time to lookup in empty hashmap
16. For element of length 1048576 it took 5.872 time to lookup in empty hashmap
17. For element of length 2097152 it took 12.003 time to lookup in empty hashmap
18. For element of length 4194304 it took 25.176 time to lookup in empty hashmap
19. For element of length 8388608 it took 50.399 time to lookup in empty hashmap
20. For element of length 16777216 it took 99.281 time to lookup in empty hashmap
Clearly the hash is O(m) where m is the length of a key.
You can make similar experiments for other mainstream languages and I expect you get a similar results.
It seems based on discussion here, that if X is the ceiling of (# of elements in table/# of bins), then a better answer is O(log(X)) assuming an efficient implementation of bin lookup.
There are two settings under which you can get O(1) worst-case times.
If your setup is static, then FKS hashing will get you worst-case O(1) guarantees. But as you indicated, your setting isn't static.
If you use Cuckoo hashing, then queries and deletes are O(1)
worst-case, but insertion is only O(1) expected. Cuckoo hashing works quite well if you have an upper bound on the total number of inserts, and set the table size to be roughly 25% larger.
Copied from here
A. The value is an int smaller than the size of the hash table. Therefore, the value is its own hash, so there is no hash table. But if there was, it would be O(1) and still be inefficient.
This is a case where you could trivially map the keys to distinct buckets, so an array seems a better choice of data structure than a hash table. Still, the inefficiencies don't grow with the table size.
(You might still use a hash table because you don't trust the ints to remain smaller than the table size as the program evolves, you want to make the code potentially reusable when that relationship doesn't hold, or you just don't want people reading/maintaining the code to have to waste mental effort understanding and maintaining the relationship).
B. You have to calculate a hash of the value. In this situation, the order is O(n) for the size of the data being looked up. The lookup might be O(1) after you do O(n) work, but that still comes out to O(n) in my eyes.
We need to distinguish between the size of the key (e.g. in bytes), and the size of the number of keys being stored in the hash table. Claims that hash tables provide O(1) operations mean that operations (insert/erase/find) don't tend to slow down further as the number of keys increases from hundreds to thousands to millions to billions (at least not if all the data is accessed/updated in equally fast storage, be that RAM or disk - cache effects may come into play but even the cost of a worst-case cache miss tends to be some constant multiple of best-case hit).
Consider a telephone book: you may have names in there that are quite long, but whether the book has 100 names, or 10 million, the average name length is going to be pretty consistent, and the worst case in history...
Guinness world record for the Longest name used by anyone ever was set by Adolph Blaine Charles David Earl Frederick Gerald Hubert Irvin John Kenneth Lloyd Martin Nero Oliver Paul Quincy Randolph Sherman Thomas Uncas Victor William Xerxes Yancy Wolfeschlegelsteinhausenbergerdorff, Senior
...wc tells me that's 215 characters - that's not a hard upper-bound to the key length, but we don't need to worry about there being massively more.
That holds for most real world hash tables: the average key length doesn't tend to grow with the number of keys in use. There are exceptions, for example a key creation routine might return strings embedding incrementing integers, but even then every time you increase the number of keys by an order of magnitude you only increase the key length by 1 character: it's not significant.
It's also possible to create a hash from a fixed-size amount of key data. For example, Microsoft's Visual C++ ships with a Standard Library implementation of std::hash<std::string> that creates a hash incorporating just ten bytes evenly spaced along the string, so if the strings only vary at other indices you get collisions (and hence in practice non O(1) behaviours on the post-collision searching side), but the time to create the hash has a hard upper bound.
And unless you have a perfect hash or a large hash table, there are probably several items per bucket. So, it devolves into a small linear search at some point anyway.
Generally true, but the awesome thing about hash tables is that the number of keys visited during those "small linear searches" is - for the separate chaining approach to collisions - a function of the hash table load factor (ratio of keys to buckets).
For example, with a load factor of 1.0 there's an average of ~1.58 to the length of those linear searches, regardless of the number of keys (see my answer here). For closed hashing it's a bit more complicated, but not much worse when the load factor isn't too high.
It is technically true because the hash function is not required to use all the information in the key and so could be constant time, and because a large enough table can bring collisions down to near constant time.
This kind of misses the point. Any kind of associative data structure ultimately has to do operations across every part of the key sometimes (inequality may sometimes be determined from just a part of the key, but equality generally requires every bit be considered). At a minimum, it can hash the key once and store the hash value, and if it uses a strong enough hash function - e.g. 64-bit MD5 - it might practically ignore even the possibility of two keys hashing to the same value (a company I worked for did exactly that for the distributed database: hash-generation time was still insignificant compared to WAN-wide network transmissions). So, there's not too much point obsessing about the cost to process the key: that's inherent in storing keys regardless of the data structure, and as said above - doesn't tend to grow worse on average with there being more keys.
As for large enough hash tables bringing collisions down, that's missing the point too. For separate chaining, you still have a constant average collision chain length at any given load factor - it's just higher when the load factor is higher, and that relationship is non-linear. The SO user Hans comments on my answer also linked above that:
average bucket length conditioned on nonempty buckets is a better measure of efficiency. It is a/(1-e^{-a}) [where a is the load factor, e is 2.71828...]
So, the load factor alone determines the average number of colliding keys you have to search through during insert/erase/find operations. For separate chaining, it doesn't just approach being constant when the load factor is low - it's always constant. For open addressing though your claim has some validity: some colliding elements are redirected to alternative buckets and can then interfere with operations on other keys, so at higher load factors (especially > .8 or .9) collision chain length gets more dramatically worse.
It is true in practice because over time it just works out as long as the hash function and table size are chosen to minimize collisions, even though that often means not using a constant time hash function.
Well, the table size should result in a sane load factor given the choice of close hashing or separate chaining, but also if the hash function is a bit weak and the keys aren't very random, having a prime number of buckets often helps reduce collisions too (hash-value % table-size then wraps around such that changes only to a high order bit or two in the hash-value still resolve to buckets spread pseudo-randomly across different parts of the hash table).
Leaving other considerations aside, the O(1) claim hinges on a constant time access model of memory, which is a good enough approximation for most practical computer science but not strictly justifiable from a theoretical point of view.
For starters, any memory addressing scheme necessarily requires multiplexing at the circuit level, which in turn requires a circuit depth at least proportional to O(log N). Since clock frequency is inversely proportional to the longest path (in number of traversed gates) of a circuit, this implies no general memory access scheme can run in less than O(log N) for fast enough CPUs or large enough memories.
Then, at a more fundamental level, you can only stack so many bits of memory within a finite distance D from the processor, and given the finite speed of light this means your worst case time for a random memory access is at least O(D^1/3), and more likely O(D^1/2) if we take into account integrated circuits are two-dimensional.
But of course in practice computers operate far from reaching these limits... or do they? This is when cache hierarchies enter the game, and why no good implementation of an algorithm or data structure can afford to ignore the actual details of the use case or the hardware implementation.
Either way the absolute worst case for a random memory access timing is given by the ping latency between your computer and some server at the opposite side of the planet, which can be in the 100s of ms and is, for the record, a lot worse than the best case scenario of having the data cached in L1 or -even better- already loaded in the registers.
As for the cost of hashing, you are correct in that it cannot be truly constant or even bounded by a set number of operations when applied to a potentially unbounded set of arbitrary-size keys such as strings, which can only be dealt with efficiently for the randomized case, but often do share arbitrarily long common prefixes that require reading and processing a number of bits larger than the size of the prefix.
For such cases it may be advisable to use a specialized data structure such as a z-fast trie or similar, which can simultaneously disambiguate prefixes and perform random memory access in amortized O(lg lg lg N).
Related
The way I understand it, an hash table is an array of linked lists, so it should actually be O(n/array_length).
Isn't saying the it's O(1) just plain wrong?
For example, if I have 1M items on hash table that is based on a 100-sized array, the average lookup would take 5,000 items. Clearly not O(1), although I assume that most implementations of a hash table use a much bigger sized array.
And what is usually the array size that is being used in most languages' (JS, Go, etc) hash table implementations?
You are correct. In general, it is not possible to say that all hash tables are O(1). It depends on some design decisions, and on other factors.
In your example, you seem to be talking about a hash table with a fixed number of buckets (100) and an unbounded number of entries N. With that design, it will take on average N / 50 comparisons to find a key that is present, and on average N / 100 comparison to discover that a key is not present. That is O(N).
There is a common implementation strategy to deal with this. As the hash table gets larger, you periodically resize the primary array, and then redistribute the keys / entries. For example, the standard Java HashMap and HashTable classes track the ratio of the array size and the number of entries. When the ratio exceeds a configurable load factor, the primary array size ts doubled. (See the javadoc for an explanation of the load factor.)
The analysis of this is complicated. However, if we can assume that keys are roughly evenly distributed across the buckets, we get the following:
average lookup times that are O(1)
average insertion times that are O(1)
worst-case insertion times that are O(N) ... when the insertion triggers a resize.
What if the key distribution is badly skewed?
This can happen if the hash function is a poor one, or if the process that generates the keys is pathological.
Well, if you don't do anything about it, the worst case occurs when all keys have the same hash value and end up in the same bucket ... irrespective of the primary array size. This results in O(N) lookup and insertion.
But there are a couple of ways to mitigate this. A simple way is to perform a second hashing operation on the key's hashcodes. This helps in some cases. A more complex way is to replace the hash chains with balanced binary trees. This changes the average behavior of lookup and insertion (for the case of pathological keys) from O(N) to O(logN)`.
From Java 8 onwards, the HashMap implementation uses either hash chains or trees, depending on the number of keys in a given bucket.
And what is usually the array size that is being used in most languages' (JS, Go, etc) hash table implementations?
For Java (and probably for the others) the array size changes over the life-time of the hash table as described above. In Java, there is an upper limit on the size or the array. Java arrays can only have 231 - 1 elements.
In the implementations you mention, the array starts small, and is reallocated with a bigger size as more elements are added.
The array size stays within a constant factor of the element count, so that the average number of elements per slot is bounded.
I don't understand how hash tables are constant time lookup, if there's a constant number of buckets. Say we have 100 buckets, and 1,000,000 elements. This is clearly O(n) lookup, and that's the point of complexity, to understand how things behave for very large values of n. Thus, a hashtable is never constant lookup, it's always O(n) lookup.
Why do people say it's O(1) lookup on average, and only O(n) for worst case?
The purpose of using a hash is to be able to index into the table directly, just like an array. In the ideal case there's only one item per bucket, and we achieve O(1) easily.
A practical hash table will have more buckets than it has elements, so that the odds of having only one element per bucket are high. If the number of elements inserted into the table gets too great, the table will be resized to increase the number of buckets.
There is always a possibility that every element will have the same hash, or that all active hashes will be assigned to the same bucket; in that case the lookup time is indeed O(n). But a good hash table implementation will be designed to minimize the chance of that occurring.
In layman terms with some hand waving:
At the one extreme, you can have a hash map that is perfectly distributed with one value per bucket. In this case, your lookup returns the value directly, and cost is 1 operation -- or on the order of one, if you like: O(1).
In the real world, implementation often arrange for that to be the case, by expanding the size of the table, etc. to meet the requirements of the data. When you have more items than buckets, you start increasing complexity.
In the worst case, you have one bucket and n items in the one bucket. In this case, it is basically like searching a list, linearly. And so if the value happens to be the last one, you need to do n comparisons, to find it. Or, on the order of n: O(n).
The latter case is pretty much always /possible/ for a given data set. That's why there has been so much study and effort put into coming up with good hashing algorithms. So, it is theoretically possible to engineer a dataset that will cause collisions. So, there is some way to end up with O(n) performance, unless the implementation tweaks other aspects ; table size, hash implementation, etc., etc.
By saying
Say we have 100 buckets, and 1,000,000 elements.
you are basically depriving the hashmap from its real power of rehashing, and also not considering the initial capacity of hashmap in accordance to need. Hashmap is more efficient in cases where each entry gets its own bucket. Lesser percentage of collision can be achieved by higher capacity of hashmap. Each collision means you need to traverse the corresponding list.
Below points should be considered for Hash table impelmentation.
A hashtable is designed such that it re sizes itself as the number of entries get larger than number of buckets by a certain threshold value. This is how we should design if we wish to implement our own custom Hash table.
A good hash function makes sure that entries are well distributed in the buckets of hashtable. This keeps the list in a bucket short.
Above takes care that access time remains constant.
I was reading the javadocs on HashSet when I came across the interesting statement:
This class offers constant time performance for the basic operations (add, remove, contains and size)
This confuses me greatly, as I don't understand how one could possibly get constant time, O(1), performance for a comparison operation. Here are my thoughts:
If this is true, then no matter how much data I'm dumping into my HashSet, I will be able to access any element in constant time. That is, if I put 1 element in my HashSet, it will take the same amount of time to find it as if I had a googolplex of elements.
However, this wouldn't be possible if I had a constant number of buckets, or a consistent hash function, since for any fixed number of buckets, the number of elements in that bucket will grow linearly (albeit slowly, if the number is big enough) with the number of elements in the set.
Then, the only way for this to work is to have a changing hash function every time you insert an element (or every few times). A simple hash function that never any collisions would satisfy this need. One toy example for strings could be: Take the ASCII value of the strings and concatenate them together (because adding could result in a conflict).
However, this hash function, and any other hash function of this sort will likely fail for large enough strings or numbers etc. The number of buckets that you can form is immediately limited by the amount of stack/heap space you have, etc. Thus, skipping locations in memory can't be allowed indefinitely, so you'll eventually have to fill in the gaps.
But if at some point there's a recalculation of the hash function, this can only be as fast as finding a polynomial which passes through N points, or O(nlogn).
Thus arrives my confusion. While I will believe that the HashSet can access elements in O(n/B) time, where B is the number of buckets it has decided to use, I don't see how a HashSet could possibly perform add or get functions in O(1) time.
Note: This post and this post both don't address the concerns I listed..
The number of buckets is dynamic, and is approximately ~2n, where n is the number of elements in the set.
Note that HashSet gives amortized and average time performance of O(1), not worst case. This means, we can suffer an O(n) operation from time to time.
So, when the bins are too packed up, we just create a new, bigger array, and copy the elements to it.
This costs n operations, and is done when number of elements in the set exceeds 2n/2=n, so it means, the average cost of this operation is bounded by n/n=1, which is a constant.
Additionally, the number of collisions a HashMap offers is also constant on average.
Assume you are adding an element x. The probability of h(x) to be filled up with one element is ~n/2n = 1/2. The probability of it being filled up with 3 elements, is ~(n/2n)^2 = 1/4 (for large values of n), and so on and so on.
This gives you an average running time of 1 + 1/2 + 1/4 + 1/8 + .... Since this sum converges to 2, it means this operation takes constant time on average.
What I know about hashed structures is that to keep a O(1) complexity for insertion removal you need to have a good hash function to avoid collisions and the structure should not be full ( if the structure is full you will have collisions).
Normally hashed structures define a kind of fill limit, by example 70%.
When the number of object make the structure be filled more than this limit than you should extend it size to stay below the limit and warranty performances. Generally you double the size of the structure when reaching the limit so that structure size grow faster than number of elements and reduce the number of resize/maintenance operations to perform
This is a kind of maintenance operation that consists on rehashing all elements contained int he structure to redistribute them in the resized structure. For sure this has a cost whose complexity is O(n) with n the number of elements stored in the structure but this cost is not integrated in the add function that will make the maintenance operation needed
I think this is what disturb you.
I learned also that the hash function generally depends on size of the structure that is used as parameter (there was something like max number of elements to reach the limit is a prime number of structure size to reduce the probability of collision or something like that) meaning that you don't change the hash function itself, you just change on of its parameters.
To answer to your comment there is not warranty if bucket 0 or 1 was filled that when you resize to 4 new elements will go inside bucket 3 and 4. Perhaps resizing to 4 make elements A and B now be in buckets 0 and 3
For sure all above is theorical and in real life you don`t have infinite memory, you can have collisions and maintenance has a cost etc so that's why you need to have an idea about the number of objects that you will store and do a trade off with available memory to try to choose an initial size of hashed structure that will limit the need to perform maintenance operations and allow you to stay in the O(1) performances
If a hash table holds N distinct items, and is not overloaded, then the hashes for the N items must have have approximately lg(N) bits, otherwise too many items will get the same hash value.
But a hash table lookup is usually said to take O(1) time on average.
It's not possible to generate lg(N) bits in O(1) time, so the standard results for the complexity of hash tables are wrong.
What's wrong with my reasoning?
What is wrong with your reasoning is the use of conflicting definitions of "time".
When one says that lookup in a hash table takes O(1) time, one usually means that it takes O(1) comparisons, that is, the number of comparisons required to find an item is bounded above by a constant. Under this idea of "time", the actual time (as in the thing you would measure in seconds) used to compute the hash causes no variation.
Measuring time in comparisons is an approximation that, while it may not reflect reality in the same way that measuring it in seconds would, still provides useful information about the behaviour of the hash table.
This sort of thing is true for most asymptotic complexity descriptions of algorithms: people often use "time" with a very abstract meaning that isn't the informal meaning of "time", but more often than not is some variation of "number of operations" (with the kind of operation often left unstated, expected to be obvious, or clear from context).
The analysis is based on the assumption that the hash function is fixed and not related to the actual number of elements stored in the table. Rather than saying that the hash function returns a lg N-bit value if there are N elements in the hash table, the analysis is based on a hash function that returns, say, a k-bit value, where k is independent of N. Typical value of k (such as 32 or 64) provide for a hash table far larger than anything you need in practice.
So in once sense, yes, a table holding N elements requires a hash function that returns O(lg n) bits; but in practice, a constant that is far larger than the anticipated maximum value of lg n is used.
Hashtable search is O(1).
I think you are mixing insertion(which is O(n)) and search.
If we look from Java perspective then we can say that hashmap lookup takes constant time. But what about internal implementation? It still would have to search through particular bucket (for which key's hashcode matched) for different matching keys.Then why do we say that hashmap lookup takes constant time? Please explain.
Under the appropriate assumptions on the hash function being used, we can say that hash table lookups take expected O(1) time (assuming you're using a standard hashing scheme like linear probing or chained hashing). This means that on average, the amount of work that a hash table does to perform a lookup is at most some constant.
Intuitively, if you have a "good" hash function, you would expect that elements would be distributed more or less evenly throughout the hash table, meaning that the number of elements in each bucket would be close to the number of elements divided by the number of buckets. If the hash table implementation keeps this number low (say, by adding more buckets every time the ratio of elements to buckets exceeds some constant), then the expected amount of work that gets done ends up being some baseline amount of work to choose which bucket should be scanned, then doing "not too much" work looking at the elements there, because on expectation there will only be a constant number of elements in that bucket.
This doesn't mean that hash tables have guaranteed O(1) behavior. In fact, in the worst case, the hashing scheme will degenerate and all elements will end up in one bucket, making lookups take time Θ(n) in the worst case. This is why it's important to design good hash functions.
For more information, you might want to read an algorithms textbook to see the formal derivation of why hash tables support lookups so efficiently. This is usually included as part of a typical university course on algorithms and data structures, and there are many good resources online.
Fun fact: there are certain types of hash tables (cuckoo hash tables, dynamic perfect hash tables) where the worst case lookup time for an element is O(1). These hash tables work by guaranteeing that each element can only be in one of a few fixed positions, with insertions sometimes scrambling around elements to try to make everything fit.
Hope this helps!
The key is in this statement in the docs:
If many mappings are to be stored in a HashMap instance, creating it with a sufficiently large capacity will allow the mappings to be stored more efficiently than letting it perform automatic rehashing as needed to grow the table.
and
The load factor is a measure of how full the hash table is allowed to get before its capacity is automatically increased. When the number of entries in the hash table exceeds the product of the load factor and the current capacity, the hash table is rehashed (that is, internal data structures are rebuilt) so that the hash table has approximately twice the number of buckets.
http://docs.oracle.com/javase/6/docs/api/java/util/HashMap.html
The internal bucket structure will actually be rebuilt if the load factor is exceeded, allowing for the amortized cost of get and put to be O(1).
Note that if the internal structure is rebuilt, that introduces a performance penalty that is likely to be O(N), so quite a few get and put may be required before the amortized cost approaches O(1) again. For that reason, plan the initial capacity and load factor appropriately, so that you neither waste space, nor trigger avoidable rebuilding of the internal structure.
Hashtables AREN'T O(1).
Via the pigeonhole principle, you cannot be better than O(log(n)) for lookup, because you need log(n) bits per item to uniquely identify n items.
Hashtables seem to be O(1) because they have a small constant factor combined with their 'n' in the O(log(n)) being increased to the point that, for many practical applications, it is independent of the number of actual items you are using. However, big O notation doesn't care about that fact, and it is a (granted, absurdly common) misuse of the notation to call hashtables O(1).
Because while you could store a million, or a billion items in a hashtable and still get the same lookup time as a single item hashtable... You lose that ability if you're taking about a nonillion or googleplex items. The fact that you will never actually be using a nonillion or googleplex items doesn't matter for big O notation.
Practically speaking, hashtable performance can be a constant factor worse than array lookup performance. Which, yes, is also O(log(n)), because you CAN'T do better.
Basically, real world computers make every array lookup for arrays of size less than their chip bit size just as bad as their biggest theoretically usable array, and as hastables are clever tricks performed on arrays, that's why you seem to get O(1)
To follow up on templatetypedef's comments as well:
The constant time implementation of a hash table could be a hashmap, with which you can implement a boolean array list that indicates whether a particular element exists in a bucket. However, if you are implementing a linked list for your hashmap, the worst case would require you going through every bucket and having to traverse through the ends of the lists.